Page:Zur Thermodynamik bewegter Systeme.djvu/4

Let $U''_{0}$ be the value now assumed by $U_{0}$; then it is

$dU=\Phi (U''_{0},\ v,\ \beta )-\Phi (U'_{0},\ v',\ \beta ),$ $d\phi =\phi (U''_{0},\ v,\ \beta )-\phi (U'_{0},\ v',\ \beta ).$

D. We bring the body to rest in an adiabatic and isochoric way. There, the work

$U''_{0}-\Phi (U''_{0},\ v,\ \beta )$

is performed. The state of the body is now given by the variables $U''_{0},\ v,\ \beta =0$. The total work of the external forces is:

$(-p_{0}dv)+(\Phi (U'_{0},\ v',\ \beta )-U'_{0})+(pdv+\beta d\phi )+(U''_{0}-\Phi (U''_{0},v,\ \beta )).$

According to the first thermodynamic main-theorem, the work must be equal to $U''_{0}-U_{0}$. The second main-theorem additionally requires, that this difference is zero. Otherwise this circular process (or the reverse one) would represent a thermal perpetual motion machine. Thus we put $U''_{0}-U_{0}$ in the previous expressions; then we notice that

${\begin{array}{lll}d\phi &=&\phi (U_{0},v,\beta )-\phi (U'_{0},v',\beta )=\\\\&=&{\frac {\partial \phi }{\partial U_{0}}}(U_{0}-U'_{0})+{\frac {\partial \phi }{\partial v}}(v-v')=\\\\&=&{\frac {\partial \phi }{\partial U_{0}}}p_{0}dv-{\frac {\partial \phi }{\partial v}}dv\end{array}}$

and

$\Phi (U'_{0},v',\beta )-\Phi (U_{0},v,\beta )={\frac {\partial \Phi }{\partial v}}dv-{\frac {\partial \Phi }{\partial U_{0}}}p_{0}dv$

thus we obtain:

${\frac {\partial \Phi }{\partial v}}dv-{\frac {\partial \Phi }{\partial U_{0}}}p_{0}dv+\beta {\frac {\partial \phi }{\partial U_{0}}}p_{0}dv-\beta {\frac {\partial \phi }{\partial v}}dv+pdv=0$

or according to (2):

$p=p_{0}{\frac {\partial H}{\partial U_{0}}}-{\frac {\partial H}{\partial v}}.$

(6)

Thus we obtain the theorem: If an arbitrary body (in the state of rest) is under the pressure $p_{0}$, then it